Centroid of a paraboloid.

Question

Consider the paraboloid given by $x^2+y^2=az$ where $a>0$ with constant density $\rho$. We are told to calculate the volume enclosed in the above paraboloid between $z=0$ and $z=h$. My attempt is $$\int_{0}^{2\pi}\int_0^h\int_0^{\sqrt{az}}rdrdzd\theta$$ which gave me $$ V=\frac{\pi ah^2}{2} $$ Which doesn't look correct based on the words of google. We continue further to derive the center of mass. $$ c = \frac{1}{V}\int_{0}^{2\pi}\int_0^h\int_0^{\sqrt{az}}r^2drdzd\theta $$ which gives the point at which the center of mass resides as $$x=y=0, z=\frac{8\sqrt{ah}}{15}$$ which certainly doesn't look right according to google. Many thanks

Answer 1

Your calculation of the volume looks fine, but the center of mass should be at $\frac{2h}{3}$:

$\displaystyle \;\;\overline{z}=\frac{1}{V}\int_0^{2\pi}\int_0^{\sqrt{ah}}\int_{r^2/a}^h z\cdot r\;dzdrd\theta=\frac{1}{V}\cdot2\pi\int_0^{\sqrt{ah}}\left[r\cdot\frac{z^2}{2}\right]_{r^2/a}^h dr$

$\displaystyle=\frac{2\pi}{V}\int_0^{\sqrt{ah}}\left(\frac{h^2}{2}r-\frac{r^5}{2a^2}\right)dr=\frac{2\pi}{V}\left[\frac{h^2r^2}{4}-\frac{r^6}{12a^2}\right]_0^{\sqrt{ah}}=\frac{2\pi}{V}\left(\frac{ah^3}{4}-\frac{ah^3}{12}\right)$

$\displaystyle=\frac{2\pi}{V}\cdot\frac{ah^3}{6}=\frac{4}{ah^2}\cdot\frac{ah^3}{6}=\frac{2h}{3}.$